72. Edit Distance 编辑距离

题目链接
tag:

• Easy;
• Dynamic Programming;

question
  Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2.

You have the following 3 operations permitted on a word:

• Insert a character;
• Delete a character;
• Replace a character;

Example 1:

Input: word1 = "horse", word2 = "ros"
Output: 3
Explanation:
horse -> rorse (replace 'h' with 'r')
rorse -> rose (remove 'r')
rose -> ros (remove 'e')

Example 2.

Input: word1 = "intention", word2 = "execution"
Output: 5
Explanation:
intention -> inention (remove 't')
inention -> enention (replace 'i' with 'e')
enention -> exention (replace 'n' with 'x')
exention -> exection (replace 'n' with 'c')
exection -> execution (insert 'u')

思路：
  这道题让求从一个字符串转变到另一个字符串需要的变换步骤，共有三种变换方式，插入一个字符，删除一个字符，和替换一个字符。根据以往的经验，对于字符串相关的题目十有八九都是用动态规划Dynamic Programming来解，这道题也不例外。这道题我们需要维护一个二维的数组dp，其中dp[i][j]表示从word1的前i个字符转换到word2的前j个字符所需要的步骤。那我们可以先给这个二维数组dp的第一行第一列赋值，这个很简单，因为第一行和第一列对应的总有一个字符串是空串，于是转换步骤完全是另一个字符串的长度。跟以往的DP题目类似，难点还是在于找出递推式，我们可以举个例子来看，比如word1是“bbc"，word2是”abcd“，那么我们可以得到dp数组如下：

Ø a b c d
Ø 0 1 2 3 4
b 1 1 1 2 3
b 2 2 1 2 3
c 3 3 2 1 2

我们通过观察可以发现，当word1[i] == word2[j]时，dp[i][j] = dp[i - 1][j - 1]，其他情况时，dp[i][j]是其左，左上，上的三个值中的最小值加1，那么可以得到递推式为：
if word1[i-1] == word2[j-1]:
  dp[i][j] = dp[i-1][j-1]
esle:
  min(dp[i-1][j-1], min(dp[i-1][j], dp[i][j-1])) + 1
代码如下：

class Solution {
public:
    int minDistance(string word1, string word2) {
        int n1 = word1.size(), n2 = word2.size();
        int dp[n1+1][n2+2];
        for (int i=0; i<=n1; ++i)
            dp[i][0] = i;
        for (int i=0; i<=n2; ++i)
            dp[0][i] = i;
        for (int i=1; i<=n1; ++i) {
            for (int j=1; j<=n2; ++j) {
                if (word1[i-1] == word2[j-1])
                    dp[i][j] = dp[i-1][j-1];
                else
                    dp[i][j] = min(dp[i-1][j], min(dp[i][j-1], dp[i-1][j-1])) + 1;
            }
        }
        return dp[n1][n2];
    }
};